Lecture 11
Acid/base equilibria.
pKa and pH
Acid/base equilibria.
1
Acid strength : the acid dissociation constant KA.
Chemistry3 Section 6.2. pp.268-270.
•
•
•
•
It is easy to quantify the
strength of strong acids since
they fully dissociate to ions in
solution.
The situation with respect to
weak acids is more complex
since they only dissociate to a
small degree in solution.
The question is how small is
small?
We quantify the idea of
incomplete dissociation of a
weak acid HA by noting that the
dissociation reaction is an
equilibrium process and
introducing the acid dissociation
constant KA.
KA values vary over a wide range
so it is best to use a log scale.
Acid dissociation
equilibrium
H3O+ (aq) + A- (aq)
HA(aq)+H2O(l)
KC 
H O A 


3
HAH 2O
K A  K C H 2O  
H O A 


3
HA
Acid dissociation
constant
KA is a measure of the acid strength.
When KA is large there is considerable
Dissociation and the acid is strong.
When KA is small there
is a small degree of dissociation, and
the acid is weak.
pK A   log10 K A
The Meaning of KA , the Acid Dissociation Constant
For the ionization of an acid, HA:
Kc =
Therefore:
Kc =
HA(aq) + H2O(l)
[H3O+] [A-]
[HA] [H2O]
[H3O+] [A-]
[HA]
Stronger acid
H3O+(aq) + A-(aq)
Since the concentration of water is
high, and does not change significantly
during the reaction, it’s value is absorbed
into the constant.
The stronger the acid, the higher the [H3O+]
at equilibrium, and the larger the Ka:
higher [H3O+]
larger Ka
For a weak acid with a relative high Ka (~10-2 ), a 1 M solution
has ~10% of the HA molecules dissociated.
For a weak acid with a moderate Ka (~10-5 ), a 1 M solution
has ~ 0.3% of the HA molecules dissociated.
For a weak acid with a relatively low Ka (~10-10 ), a 1 M solution
has ~ 0.001% of the HA molecules dissociated.
2
Acid strengths and
Ka value.
The Relationship Between Ka and pKa
Acid Name (Formula)
pK A   log10 K A
pKA
KA at 298 K
Hydrogen sulfate ion (HSO4-)
1.02 x 10-2
1.991
Nitrous acid (HNO2)
7.1 x 10-4
Acetic acid (CH3COOH)
1.8 x 10-5
Hypobromous acid (HBrO)
2.3 x 10-9
8.64
Phenol (C6H5OH)
1.0 x 10-10
10.00
KA 
pK A 
3.15
4.74
When KA is small pKA is large and the acid does not dissociate in solution
to a large extent. A change in 1 pKA unit implies a 10 fold change in KA value
and hence acid strength.
3
Ka and pKa
Acid
Strength
decreases
pK A   log10 K A
pKa calculation
Ka=
[CH3CO2-][H3O+]
= 1.8x10-5
[CH3CO2H]
pKa= -log(1.8x10-5) = 4.74
4
Weak acids
Acid-Base Properties of Water
H+ (aq) + OH- (aq)
H2O (l)
autoionization of water
H
O
H
+ H
[H
O
H
]
H
+
+ H
O
-
H
base
H2O + H2O
acid
O
conjugate
acid
H3O+ + OHconjugate
base
15.2
5
Autoionization of Water
H2O(l) + H2O(l)
Kc =
H3O+ + OH-
[H3O+][OH-]
[H2O]2
The ion-product for water, Kw:
Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0 x 10-14 (at 298K)
For pure water the concentration of hydroxyl and hydronium ions
must be equal:
[H3O+] = [OH-] =
1.0 x 10-14 = 1.0 x 10 -7 M (at 25°C)
The molarity of pure water is:
1000g/L
18.02 g/mol
= 55.5 M
The Ion Product of Water
H2O (l)
H+ (aq) + OH- (aq)
Kc =
[H+][OH-]
[H2O]
[H2O] = constant
Kc[H2O] = Kw = [H+][OH-]
The ion-product constant (Kw) is the product of the molar
concentrations of H+ and OH- ions at a particular temperature.
At
Kw =
250C
[H+][OH-]
= 1.0 x
10-14
[H+] = [OH-]
Solution Is
neutral
[H+] > [OH-]
acidic
[H+] < [OH-]
basic
15.2
6
Weak Bases and Base Ionization Constants
NH4+ (aq) + OH- (aq)
NH3 (aq) + H2O (l)
Kb =
[NH4+][OH-]
[NH3]
Kb
weak base
strength
Kb = base ionization constant
7
Basicity Constant Kb.
BH+ (aq) + OH- (aq)
B(aq) + H2O (l)
• The proton accepting
strength of a base is
quantified in terms of the
basicity constant Kb.
• The larger the value of Kb,
the stronger the base.
• If Kb is large then pKb will be
small, and the stronger will
be the base.
• Solve weak base problems
like weak acids except solve
for [OH-] instead of [H+].
KC 
BH OH 


BH 2O
K b  K C H 2O  
BH OH 


B
pK b   log10 K b
K a K b  KW
pK a  pK b  pKW
Ionization Constants of Conjugate Acid-Base Pairs
H+ (aq) + A- (aq)
HA (aq)
A- (aq) + H2O (l)
Ka
OH- (aq) + HA (aq)
H+ (aq) + OH- (aq)
H2O (l)
Kb
Kw
KaKb = Kw
Weak Acid and Its Conjugate Base
Ka =
Kw
Kb
Kb =
Kw
Ka
15.7
8
Acid and base dissociation constants
Determining a Value of KA from the pH of a Solution of
a Weak Acid.
Butyric acid, HC4H7O2 (or CH3CH2CH2CO2H) is used to
make compounds employed in artificial flavorings and
syrups. A 0.250 M aqueous solution of HC4H7O2 is found to
have a pH of 2.72. Determine KA for butyric acid.
HC4H7O2 + H2O  C4H7O2 + H3O+
Ka = ?
For HC4H7O2 KA is likely to be much larger than KW.
Therefore assume self-ionization of water is unimportant.
HC4H7O2 + H2O  C4H7O2 + H3O+
Initial conc.
Changes
Equilbrm. conc.
0.250 M
-x M
(0.250-x) M
0
+x M
xM
0
+x M
xM
9
Log[H3O+] = -pH = -2.72
[H3O+] = 10-2.72 = 1.9x10-3 = x
Ka=
[H3O+] [C4H7O2-]
[HC4H7O2]
Ka= 1.5x10-5
SAMPLE PROBLEM 18.7:
PROBLEM:
PLAN:
1.9x10-3 · 1.9x10-3
=
(0.250 – 1.9x10-3)
Check assumption: Ka >> KW.
Determining Concentrations from Ka and
Initial [HA]
Propanoic acid (CH3CH2COOH, which we simplify and HPr) is
an organic acid whose salts are used to retard mold growth in
foods. What is the [H3O+] of 0.10M HPr (Ka = 1.3x10-5)?
Write out the dissociation equation and expression; make whatever
assumptions about concentration which are necessary; substitute.
Assumptions:
For HPr(aq) + H2O(l)
H3O+(aq) + Pr-(aq)
x = [HPr]diss = [H3O+]from HPr= [Pr-]
Ka =
[H3O+][Pr-]
[HPr]
SOLUTION:
Concentration(M)
Initial
Change
Equilibrium
HPr(aq) + H2O(l)
H3O+(aq) + Pr-(aq)
0.10
-
0
0
-x
-
+x
+x
0.10-x
-
x
x
Since Ka is small, we will assume that x << 0.10
10
SAMPLE PROBLEM 18.7:
Determining Concentrations from Ka and
Initial [HA]
continued
1.3x10-5 =
[H3O+][Pr-]
[HPr]
x  (0.10)(1.3x105 )
(x)(x)
=
0.10
= 1.1x10-3M = [H3O+]
Check: [HPr]diss = 1.1x10-3M/0.10M x 100 = 1.1%
The relationship between [H3O+] and [OH-] and the
relative acidity of solutions.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
[H3O+]
Divide into Kw
[OH-]
[H3O+] > [OH-]
[H3O+] = [OH-]
[H3O+] < [OH-]
ACIDIC
SOLUTION
NEUTRAL
SOLUTION
BASIC
SOLUTION
11
The pH concept.
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The best quantitative measure
of acidity or alkalinity rests in
the determination of the
concentration of hydrated
protons [H3O+] present in a
solution.
The [H3O+] varies in magnitude
over quite a large range in
aqueous solution, typically from 1
M to 10-14 M.
Hence to make the numbers
meaningful [H3O+] is expressed
in terms of a logarithmic scale
called the pH scale.
The higher the [H3O+] , the
more acidic the solution and the
lower is the solution pH.

pH   log10 H 3O 
H O   10

Linear and logarithmic
Scales.

 pH
3
Strong acids and bases
HCl
CH3CO2H
Thymol Blue Indicator
pH < 1.2 < pH < 2.8 < pH
12
The pH Scale.
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pH is expressed on a numerical scale from 0 to 14.
When [H3O+] = 1.0 M (i.e. 100M), pH = 0.
When [H3O+] = 10-14 M, pH = 14.
pH value < 7 implies an acidic solution.
pH value > 7 implies an alkaline solution.
pH value = 7 implies that the solution is neutral.
The definition of pH involves logarithms. Hence a
change in one pH unit represents a change in
concentration of H3O+ ions by a factor of 10.
[H3O+]
1.0 M
10-7 M
0
7
10-14 M
pH
14
The pH Values of
Some Familiar Aqueous
Solutions

pH   log10 H 3O 

pOH   log10 OH



13
pH – A Measure of Acidity
pH = -log [H+]
Solution Is
neutral
[H+] = [OH-]
At 250C
[H+] = 1 x 10-7
pH = 7
acidic
[H+] > [OH-]
[H+] > 1 x 10-7
pH < 7
basic
[H+] < [OH-]
[H+] < 1 x 10-7
pH > 7
pH
[H+]
Calculating [H3O+], pH, [OH-], and pOH
PROBLEM: In a restoration project, a conservator prepares copper-plate
etching solutions by diluting concentrated HNO3 to 2.0M, 0.30M,
and 0.0063M HNO3. Calculate [H3O+], pH, [OH-], and pOH of
the three solutions at 250C.
PLAN:
HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the [OH-]
and then convert to pH and pOH.
SOLUTION:
For 2.0M HNO3, [H3O+] = 2.0M and -log [H3O+] = -0.30 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/2.0 = 5.0x10-15M; pOH = 14.30
For 0.3M HNO3, [H3O+] = 0.30M and -log [H3O+] = 0.52 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/0.30 = 3.3x10-14M; pOH = 13.48
For 0.0063M HNO3, [H3O+] = 0.0063M and -log [H3O+] = 2.20 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/6.3x10-3 = 1.6x10-12M; pOH = 11.80
14
The problem of acid rain.
Methods for Measuring the pH of an
Aqueous Solution
(a) pH paper
(b) Electrodes of a pH meter
15
pH Measurement.
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•
Approximate pH of a solution
determined by use of acid/base
indicators.
–
–
–
–
Indicators are substances (weak
acids) which change colour over a
specific pH range when they donate
protons.
We add a few drops of indicator
(which changes colour over the
required pH range) to the test
solution and record the colour
change produced.
This procedure is utilized in
acid/base titrations. Universal
indicator (mixture of pH indicators)
often used for making approximate
pH measurements in range 3-10.
As solution pH increases, the
indicator changes colour from red to
orange to yellow to green to blue, and
finally to purple.
HIn(aq )  H 2O  H 3O  (aq )  In 
More accurate pH values determined
using an electronic instrument called
a pH meter.
–
–
–
–
–
The device (consisting of a probe
electrode made of glass and
associated electronics) measures the
electrical potential generated across
a glass membrane (which separates
an internal solution of known [H3O+]
from the external test solution of
unknown [H3O+]) located at the
electrode tip.
This membrane potential is
proportional to the pH of the test
solution.
A digital readout of solution pH is
obtained.
The pH meter is essentially a
voltmeter connected to a chemical
sensor probe which is sensitive to
the concentration of hydrated
protons.
The pH meter is an example of a
potentiometric chemical sensor
system. In a potentiometric chemical
sensor, the measured voltage is
proportional to the logarithm of the
analyte concentration.
16
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